EXPLANATION OF COPPER IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 26 , 2015 Copper is a chemical element with symbol Cu and atomic number 29. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light (FASTER THAN LIGHT), which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Copper including the following ground state electron configuration 1s22s22p63s23p63d104s1. This is in the group of elements known as the transitional metals. As we progress through this group the 3d sub-shell is gradually filled, The electronic configuration of transition metal elements are characterized as having full outer sub-orbitals and the second outermost d sub-orbitals incompletely filled, with the exception of Copper which loses one 4s orbital electron to the 3d sub-orbital for increased stability. So we observe the following actual electron configuration: The loss of the electron from the 4s orbital completes the 3d sub-orbital and leaves it in a more stable, lower energy state. According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies of copper (in eV) are the following: E1 = 7.73, E2 = 20.29, E3 = 36.84 , E4 = 57.38 , E5 = 79.8, E6 = 103, E7 = 139, E8= 166, E9 = 199, E10 = 232, E11 = 265.3 , E12 = 369, E13 = 401, E14 = 435 , E15 = 484 , E16 = 520, E17 = 557, E18 = 633, E19 = 670.6, E20 = 1697, E21 = 1804, E22 = 1916, E23 = 2060 , E24 = 2182, E25 = 2308 , E26 = 2478 , E27 = 2587.5, E28 = 11062.38, and E29 = 11567.62 . Firstly we examine the - E1 = E(4s1). Here the E(4s1) represents the binding energy of the one outermost electron. Secondly, we observe that the - ( Ε2 + E3 + E4 + E5 + Ε6 + Ε7 + E8 + E9 + E10 + Ε11 ) = E(3d10). The E(3d10) represents the binding energy of 10 paired electrons (3d10) of five orbitals. Then, the - ( E12 + E13 + E14 + E15 + E16 +E17 ) equals the binding energy E(3p6) of the 6 paired electrons. Whereas the - ( E18 + E19) equals the binding energy E( 3s2). Also, the - ( E20 + E21 + E22 + E23 + E24 + E25 ) equals the binding energy E(2p6) . On the other hand the -( E26 + E27 ) equals the binding energy E(2s2) , while the - ( E28 + E29 ) equals the binding energy E(1s2). For understanding better the ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF E1 = 7.73 eV = -E(4s1) Here the binding energy E(4s1) of the one outermost electron is given by applying the Bohr formula. The charges (-28e) of the inner electrons (1s22s22p63s23p63d10) screen the nuclear charge (+29e) and for a perfect screening we would have an effective ζ = 1. However according to the quantum mechanics the electron 4s1 penetrates the ( 3d10) leading to ζ > 1. Under this condition we may write E1 = 7.73 eV = -E(4s1) = - (13.6057 )ζ2 / n2 Since n = 4 we get ζ = 3 > 1 . Here the 3 > 1 means that the one outermost electron (4s1) leads to the deformation of 3d10 . ' ' EXPLANATION OF - ( E2 +…. + E11 ) = - 1298.6 = E(3d10) ' '''Here the E(3d10) represents the binding energy of the 10 electrons (3d10) . Note that here the 10 electrons create five orbitals with electrons of opposite spin given by applying my formula of 2008. The charges (-18e) of the inner electrons (1s22s22p63s23p6) screen the nuclear charge (+29e) and for a perfect screening we would have ζ = 11. Under this condition we may write ' E2 +..+ E11 ) = 1298.6 eV = - E(3d10) = -5+ (16.95)ζ - 4.1 / n2 So using n = 3 we may rewrite 15.1167ζ2- 9.4167ζ - 1298.144 = 0 Surprisingly solving for ζ we get ζ = 9.58 < 11 , which cannot exist . In fact, the 10 electrons of the five orbitals make a complete spherical sub- shell leading to a perfect screening with ζ = 11. Thus using ζ = 11 we expect to determine n > 3. Under this condition we may write E2 +..+ E11 ) = 1298.6 eV = -E(3d10) = -527.21)112 - (16.95)11 + (4.1) / n2 Then solving for n we get n = 3.46 > 3 . ' ' '''EXPLANATION OF ( E12 + … + E17 ) = 2766 eV = -E(3p6) Here the E(3p6) represents the binding energy of the 6 paired electrons given by applying my formula of 2008. ' '''The charges (-12e) of the twelve inner electrons like (1s22s22p63s2 ) screen the nuclear charge (+29e) and for a perfect screening we would have an effective Zeff = ζ = 17. However the 6 paired electrons ( 3p6 ) repel the 3s2 electrons and lead to the deformation of shells with ζ > 17. Under this condition we may write ( E12 +…+ E17) = 2766 eV' = '-E(3p6)' = '-3+(16.95)ζ - 4.1 / n2 Since n = 3 the above equation can be written as 9.07ζ2 - 5.65ζ - 2764.63 = 0 Then, solving for ζ we get ζ = 17.773 > 17 . ' ' '''EXPLANATION OF ( E18 + E19 ) = 1303.6 eV = -E(3s2) ' Here the E(3s2) represents the binding energy of the two paired electrons (3s2) given by applying my formula of 2008. The charges (-10e) of the inner 10 electrons of 1s2.2s2.2p6 screen the nuclear charge (+29e) and for a perfect screening we would have an effective ζ = 19. However the two electrons of 3s2 penetrate the 2p6 leading to the deformation of shells with ζ > 19. Under this condition we write the following equation as ( E18 + E19 ) = 1303.6 eV = - E(3s2) = - 27.21)ζ2 + (16.95) ζ - 4.1 / n2 Since n = 3 we may write 3.0233ζ2 - 1.8833ζ - 1303.1444 = 0 Then solving for ζ we get ζ = 21 > 19 . ' ' EXPLANATION OF (E20 + …+ E25 ) = 11967 eV = -E(2p6) ''' Here E(2p6) represents the binding energy of the six paired electrons. The charges (-4e) of the four inner electrons (1s22s2) screen the nuclear charge (+29e) and for a perfect screening we would have ζ = 25. Thus for n = 2 we should determine the effective ζ = 25 of the total binding energy of 6 paired electrons by applying my formula of 2008 as ( E20 + …+ E25) = 11967 eV = -E(2p6) = -3+ (16.95)ζ - 4.1) / n2 Surprisingly using n = 2 and solving for ζ we get ζ < 25 , which cannot exist. In fact, the 2px2 , 2py2 , and 2pz2 exert electrical repulsions from symmetrical positions and make a complete spherical shell. So they provide a perfect screening with ζ = 25. Thus using ζ = 25 and solving for n we expect to find n > 2, because a perfect screening after the experiments of ionizations means that the quantum number n = 2 becomes n > 2. Under this condition for determining here the quantum number n the above equation could be written as ( E20 +…+ E25 ) = 11967 eV = -E(2p6) = -3+ (16.95)25 - 4.1 ) / n2 Then solving for n we get n = 2.039 > 2 In other words the three orbitals of paired electrons do not lead to the deformations of 1s2 and 2s2 but differ from the symmetry of (2px1 + 2py1 + 2pz1) which exert both electric and magnetic repulsions. Here the electric repulsions between the paired electrons of 2px2, 2py2 and 2pz2 make a complete spherical shell and lead to a perfect screening with ζ = 25. Under this condition the quantum number n = 2 becomes n = 2.039 . Note that the two electrons of opposite spin (say the 2px2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of the 2px2 or 2py2 or 2pz2 , today many physicists believe incorrectly that it is due to the Coulomb repulsion between the two electrons of opposite spin. Under such fallacious ideas I published my paper of 2008. ' ' '''EXPLANATION OF ( E26 + E27 ) = 5065.5 eV = - E(2s2) Here the E(2s2) represents the binding energy of the two paired electrons (2s2). The charges (-2e) of 1s2 screen the nuclear charge (+29e) and for a perfect screening we would have an effective ζ = 27. However according to the quantum mechanics the two electrons (2s2) penetrate the 1s2 shell. Thus they lead to the deformations of both 1s2 and 2s2 spherical shells giving an effective ζ > 27. Since n = 2 we apply my formula of 2008 to write ( E26 + E27 ) = - E(2s2) = - )ζ2 + ( 16.95)ζ - 4.1 / 22 Since ( E26 + E27 ) = 5065.5 eV, we may rewrite 6.8025ζ2 - 4.2375ζ - 5064.475 = 0 Then, solving for ζ we get ζ = 27.59 > 27 . Here ζ = 27.59 > 27 means that the repulsions (2s2-1s2 ) lead to the deformation of shells, because the two electrons (2s2) or (1s2) of opposite spin behave like one particle. Note that in both cases the repulsions are due to only electric forces of the Coulomb law. Whereas in the case of the three electrons of 2px1, 2py1, and 2pz1 of parallel spin (S = 1) the three electrons interact with both electric and magnetic repulsions from symmetrical positions. 'EXPLANATION OF - ( Ε28 + E29 ) = - 22630 = E( 1s2) ' As in the case of helium the binding energy E(1s2) is due to the two remaining electrons of 1s2 with n = 1. Thus we may calculate the binding energy by applying my formula of 2008 for Z = 29 as E(1s2) = )292 + (16.95)29 - 4.1 /12 = - 22396.16 However the experiments of ionizations give - (E28 + E29 ) = - 22630 . ''' '''In other words my formula of 2008 gives the value of 22396.16 eV, which is smaller than the experimental value of 22630 eV. Under this condition of ionizations I suggest that n = 1 becomes n < 1 due to the fact that the ionizations reduce the electron charges and now the nuclear charge is much greater than the electron charge of the two remaining electrons. So for Z = 29 we determine the n by writing (E28 + E29 ) = 22630 eV = - E(1s2) = - )292 + (16.95)29 - 4.1 /n2 Then solving for n we get n = 0.9948. Category:Fundamental physics concepts